Question

The product of three consecutive numbers in G.P. is 27 and the sum of the products of numbers taken in pair is 39. Find the numbers.

Answer 1

Let `a/r`, a, ar

`a/r xx a xx ar` = 27

a³ = 27

a³ = 3³

a = 3

`a/r(a) + a(ar) + a/r(ar)` = 39

`3/r(3) + 3(3r) + 3(3)` = 39

`9[1/r + r + 1]` = 39

`1/r + r + 1 = 39/9`

`1 + r + r^2 = (13r)/3`

3(1 + r + r²) = 13r

3 + 3r + 3r² = 13r

3r² + 3r − 13r + 3 = 0

3r² − 9r − r + 3 = 0

3r(r − 3) − 1(r − 3) = 0

(r − 3) (3r − 1) = 0

r = 3, `1/3`

If a = 3, r = 3

⇒ `a/r`, a, ar

⇒ `3/3`, 3, 3(3)

⇒ 1, 3, 9

If a = 3, r = `1/3`

⇒ `3 xx 3, 3, 3(1/3)`

⇒ 9, 3, 1